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Rings and Polynomials and other voodoo

  1. Jan 14, 2015 #1
    1. The problem statement, all variables and given/known data
    Prove the following theorem: Let R be any ring and let f =! 0 and g =! 0 (they don't equal zero) be polynomials in R[x]. If the leading coefficient of either f or g is a unit in R, then:
    1) fg =! 0 in R[x]
    2) deg(fg) = deg(f) + deg(g)

    2. Relevant equations


    3. The attempt at a solution
    So i set up two polynomials f and g, where |g| = n and |f| = m. I multiplied these together and showed that the last term is equal to x^(mn) multiplied by the summation of dual combinations of coefficients such that i+j = mn where i represents the i'th coefficient on in f and j represents the j-th coefficient in g. I'm trying to show that this summation does not equal zero, and therefore deg(fg) = mn = deg(f) + deg(g), which would also show that fg =! 0 in R[x]

    Anyway, that's how far I am. I'm trying to figure out what the lead coefficient of f being a unit has to do with it... any advice would be koo koo kachu.
     
  2. jcsd
  3. Jan 14, 2015 #2

    LCKurtz

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    Do you mean ##x^{m+n}##?

    Again, I think you mean ##i+j = m+n##. Have you written out what that summation looks like as the coefficient of ##x^{m+n}##? Try it for the ring being the reals.
     
  4. Jan 14, 2015 #3
    yes, my bad, m+n. Yeah, it works for the ring being the reals and that much is pretty obvious really. I'm trying to figure out though how the lead coefficient of f being a unit guarantee's that deg(fg) = deg(f) + deg(g). I understand if the ring is the reals then the lead coefficient of both f and g will be zero, and since the real's are also a field and hence an integral domain it becomes quite a bit more trivial to prove it seems. So yeah, the real gives too many added bonus's that make it easy to see, but unfortunately our question doesn't say that our ring R is a field, or even a division ring for that matter.
     
  5. Jan 14, 2015 #4

    LCKurtz

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    Huh?? What is the coefficient of ##x^{m+n}##?
     
  6. Jan 14, 2015 #5

    Stephen Tashi

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    Are premises stated correctly? If the ring is [itex] \mathbb{Z}/(\mathbb{4Z}) [/itex] , "mod 4 arithmetic" , what is the product of [itex] (1 + x + 2 x^2)( 1 + 2x^2) ? [/itex]
     
  7. Jan 14, 2015 #6

    Dick

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    Neither leading coefficient is a unit in [itex] \mathbb{Z}/(\mathbb{4Z}) [/itex], so that's where it violates the premises.
     
  8. Jan 14, 2015 #7

    Stephen Tashi

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    I had wong idea of what "leading" coefficent meant! So, PsychonautQQ should look through his course materials for theorems about the units of ring.
     
  9. Jan 14, 2015 #8

    jbunniii

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    The leading coefficient is the coefficient of the term with the highest power. So (except for the zero polynomial), the leading coefficient can't be zero. Example: the leading coefficient of ##5x^3 + 4x^2 + 2x - 1## is ##5##.

    In general, if ##f## is a polynomial with degree ##m## then it is of the form ##f(x) = a_m x^m + a_{m-1}x^{m-1} + \cdots + a_1 x + a_0##, where ##a_m## is the leading coefficient of ##f##. Note that ##a_m## is not zero, otherwise the degree of ##f## would be less than ##m##.

    Similarly, if ##g## has degree ##n## then it looks like ##g(x) = b_n x^n + b_{n-1}x^{n-1} + \cdots + b_1 x + b_0##, where ##b_n \neq 0## is the leading coefficient of ##g##.

    Now multiply these polynomials together. The highest power of ##x## which will appear in the result is ##x^{m+n}##. What is the coefficient of that term?
     
  10. Jan 15, 2015 #9

    Dick

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    In a general ring, it's not so important that both are not zero, though that's a good point. It's more important that one is a unit. But I'm sure you were getting to that.
     
  11. Jan 15, 2015 #10

    jbunniii

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    Right, I just wanted to address his remark that the leading coefficient of a polynomial could be zero. It suggested a misunderstanding of what "leading coefficient" means, which would make it hard to solve the problem.
     
  12. Jan 15, 2015 #11

    LCKurtz

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    And you'll notice, PsychonautQQ hasn't replied to post #4 yet.

    @PsychonautQQ Are you going to update us on your progress on this thread or just abandon it?
     
    Last edited: Jan 17, 2015
  13. Jan 19, 2015 #12
    I'm sorry, i meant obviously if the ring's are over the reals then the lead coefficient of f*g will NOT be zero! The coefficient of x^(m+n) will be the lead coefficient of f multiplied by the lead coefficient of g. However, the reason I am so sure of this is because the reals are a field and thus an integral domain. The stated question isn't so generous, the only information it gives is that the lead coefficient of ONE of the polynomials is a unit.

    Have I stated my dilemma clearly? Sorry, i'm not good at speaking clearly in mathematical terms.
     
  14. Jan 19, 2015 #13

    Stephen Tashi

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    If [itex] u [/itex] is a unit in the ring and [itex] r [/itex] is not zero, can [itex] ur = 0 [/itex] ?

    Consider [itex] u^{-1} (ur) [/itex].
     
  15. Jan 19, 2015 #14
    Thank you. I understand now.
     
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